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Determine functions u and v satisfying given conditions

by RoRi

The graph of a solution u of the differential equation

    \[ y'' - 3y' - 4y = 0 \]

intersects the graph of a solution v of the differential equation

    \[ y'' + 4y' - 5y = 0 \]

at the origin. Determine formulas for the functions u and v if the curves have equal slopes at the origin and if

    \[ \lim_{x \to +\infty} \frac{(v(x))^4}{u(x)} = \frac{5}{6}. \]

First, we compute the general solutions of the two second-order differential equations.

    \[ y'' - 3y' - 4y = 0 \]

is of the form

    \[ y'' + ay' + by = 0 \qquad \text{with} \qquad a = -3, \ \ b = -4. \]

This gives us d = a^2 - 4b = 25 so d > 0 and k = \frac{1}{2} \sqrt{d} = \frac{5}{2}. Hence, the general solutions are

    \begin{align*} y &= e^{-\frac{ax}{2}} \left( c_1 e^{kx} + c_2 e^{-kx} \right) \\ &= e^{\frac{3x}{2}} \left( c_1 e^{\frac{5}{2}x} + c_2 e^{-\frac{5}{2}x} \right) \\ &= c_1 e^{4x} + c_2 e^{-x}. \end{align*}

The other equation

    \[ y'' + 4y' - 5y = 0 \]

is of the form

    \[ y'' + ay' + by = 0 \qquad \text{with} \qquad a = 4, \ \ b = -5. \]

Therefore, d = a^2 -4b = 36 so k = \frac{1}{2} \sqrt{d} = 3 and the general solutions are given by

    \begin{align*} y &= e^{-\frac{ax}{2}} \left( c_1 e^{kx} + c_2 e^{-kx} \right) \\ &= e^{-2x} \left( c_1 e^{3x} + c_2 e^{-3x} \right) \\ &= c_1 e^x + c_2 e^{-5x}. \end{align*}

Letting u(x) and v(x) denote the particular solutions we are looking for (and renaming the constants a_1, \ a_2 and b_1, \ b_2 in the equations for u and v) we have

    \begin{align*} u(x) &= a_1 e^{4x} + a_2 e^{-x} & \implies && u'(x) &= 4a_1 e^{4x} - a_2 e^{-x} \\ v(x) &= b_1 e^x + b_2 e^{-5x} & \implies && v'(x) &= b_1 e^x - 5b_2 e^{-5x}. \end{align*}

We are given u(0) = v(0) = 0 and u'(0)= v'(0). So,

    \[ u(0)= v(0) = 0 \quad \implies \quad a_1 + a_2 = b_1 + b_2 0 \]

and

    \[ v'(0) = u'(0) \quad \implies \quad b_1 - 5b_2 = 4a_1 -a_2. \]

Finally, we are also given

    \[ \lim_{x \to +\infty} \frac{(b_1 e^x + b_2 e^{-5x})^4}{a_1 e^{4x} + a_2 e^{-x}} = \frac{5}{6} \quad \implies \quad \frac{b_1^4}{a_1} = \frac{5}{6}. \]

Putting these all together and solving for the constants we obtain

    \[ b_1 = 1, \quad b_2 = -1, \quad a_1 = \frac{6}{5}, \quad a_2 = -\frac{6}{5}. \]

Therefore, the functions we are looking for are

    \[ u(x) = \frac{6}{5} \left( e^{4x} - e^{-x} \right), \quad v(x) = e^x - e^{-5x}. \]

4 comments

    • S says:

      Actually it doesn’t satisy the limit, so the solution I proposed here is incorrect, and the one here and in the book is the correct one.

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